What is the voltage drop across a capacitor?

The voltage drop across a capacitor is proportional to its charge, and it is uncharged at the beginning; whereas the voltage across the resistor is proportinal to the current and there is a current at the start. With less current, the rate at which charge goes onto the capacitor decreases.

Why does voltage decrease when discharging a capacitor?

Correct, as you discharge a capacitor the voltage drops. This is due to the relationship of Q=VC – the charge stored in a capacitor is proportional to the voltage for a given capacitance. As you discharge the capacitor, the charge on the capacitor is reduced, and so the voltage reduces.

Why does a capacitor discharge exponentially?

When capacitor is charged it acts like a battery but the moment any load is connected it starts discharging i.e. the opposite charges start moving in such a direction so that they neutralize each other as quickly as possible. The fastest possible discharging path can be exponential.

Why does voltage drop during discharge?

The voltage also drops during discharge as the cell resistance increases due to the accumulation of discharge products, activation and concentration, polarization, and related factors.

Does voltage decrease exponentially?

As the stored energy decreases, the voltage across decreases which (again by Ohm’s law), means the current through decreases and so the rate at which the energy decreases is also decreasing. This leads to the exponential decrease in voltage across the capacitor.

What is the voltage across a capacitor at steady state?

The circuit is at steady state when the voltage and the current reach their final values and stop changing. In steady state, the capacitor has a voltage across it, but no current flows through the circuit: the capacitor acts like an open circuit.

What is the voltage across the capacitor when the source voltage is 100V and the Q factor is 10?

1000V
What is the voltage across the capacitor when the source voltage is 100V and the Q factor is 10. Solution: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage =10*100=1000V.